Past Year Exam Q CH3

8/11/2019 Past Year Exam Q CH3

1/13

Case 2

Chew T S MA1506-14 Chapter 3 1

2

2

2

4( )( ) 14

2( ) 2 2

B B s E B BB sE

s s s s

2( )F N sN BN E

2

4

BE

s

where1 2

B

s

2

2

1

4( )( ) 14

2( ) 2 2

B B s E BB sE

s s s

3.4 Harvesting

Warning:1 2

2 1NOT


2/13

2

1 2( ) ( )( )( )

dNF N s N N

dt

1

2

2( )F N sN BN E

0

dN

dt 0

dN

dt 0

dN

dt

N

3.4 Harvesting

22 10 12 ( 2 6)( 2) ( 2)( 3)( 2)N N N N N N

2 22 10 12 ( 2)( 5 6) ( 2)( 3)( 2)N N N N N N

WHY?

See the last

two examples


3/13

Chew T S MA1506-14 Chapter 3 3

N

t

1

2

equi soln

stable

equi soln

unstable

2

4

BE

s

3.4 Harvesting

B/s -

Good region

able to bounce back

Region 1

Region 2

Region 3

We have to ensure that the # of fish is not in the region 1,

since in this case , eventually, the population of fish will be zero.


4/13

/B s

1

2

2

1.512

dN NN

dt

Initial population

under harvesting modelis B/s


5/13

2( ) ( )dN

B D N B sN N BN sNdt

2dN BN sN Edt

1

2

B

s

2


6/13

Chew T S MA1506-14 Chapter 3

3.4 Harvesting

6

1 2 2

2

0.2

0.8

B/S

2

1

2

4

BE

s


7/13

2

0ax bx c

Recall

Suppose that one of the root is known, and a, b

are also given. How to find the constant c

It is known that

2

1 2

b

a

Hence we can find1

Furthermore, 1 2ca

Hence we can find c

Certainly, we can use formula

2

1 2

4,

2

b b ac

a

to find the value c.

But the computation

of the 1st

method is easier

2

1 2[ ( / ) ( / )] ( )( )a x b a x c a a x x


8/13

2100000 ( 100000) 4(1)(100000 )68002

E

0.1B 0.1B


9/13

NOT a harvesting model

1 2 0.0116 40.0029

ba

1 2/ / ( 0.0029) ( 2)(6)c a E

2 6

1 2

6

2

( 0.0029)(N 2)(N 6)dN

dt


10/13

2012

10

20

3


11/13

05

10

1sinh ( )

2

x xx e e

sinh 0 0x iff x

sinh 0 iff 0x x sinh 0 iff 0x x

(20 2 )(5 )

2 (10 )(5 )

similar to

dxx x x

dt

x x x


12/13


13/13

Past Year Exam Q CH3

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